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Mathematics

1 answer:

VARVARA [1.3K]2 years ago

7 0

Answer:

11 and 6

1 and -8

1 and -60

4 and 2

6 and -12

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2x-9<-1 pleassssssse help me

e-lub [12.9K]

Answer:

x <4

Step-by-step explanation:

2x-9<-1

Add 9 to each side

2x-9+9<-1+9

2x < 8

Divide each side by 2

2x/2 < 8/2

x <4

5 0

2 years ago

PLEASE HELP!!!!!!! What is the mean absolute deviation (MAD) of the data set?

melisa1 [442]

Answer: 2.8

Step-by-step explanation:

The mean of the data set it 10. Then find the difference between the mean and each data value. |10-5|=5 do this for all of them. You get 5,1,1,1,6. Now find the mean of the these new data value. 5+1+1+1+6=14 14/5=2.8

7 0

2 years ago

Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}$